SOLVING TRIGONOMETRIC EQUATIONS

Note: If you would like a review of trigonometry, click on
trigonometry.

Example 1: Solve for x in the following equation.

There are an infinite number of solutions to this problem.

We can make the solution easier if we convert all the trigonometric terms to
cosine.

One common trigonometric identity is
and
Replace the
.
in the original equation with
and we have an equivalent equation with cosines terms.

Isolate the cosine term. To do this, rewrite the left side of the equation
in an equivalent factored form.

The product of two factors equals zero if at least one of the factors equals
zero. This means that
if
or

We just transformed a difficult problem into two easier problems. To find
the solutions to the original equation,
,
we find the solutions to the equations
and

and

How do we isolate the x? We could take the arccosine of both sides. However,
the cosine function is not a one-to-one function.

Let's restrict the domain so the function is one-to-one on the restricted
domain while preserving the original range. The graph of the cosine function
is one-to-one on the interval
If we restrict the
domain of the cosine function to that interval , we can take the arccosine
of both sides of each equation.

We know that
Therefore, if
,
then

We complete the problem by solving for the second factor.

As stated previously, we know that
Therefore, if
,
then

Since the period of
equals ,
these solutions will repeat
every
units. The exact solutions are

where n is an integer.

The approximate values of these solutions are

where n is an integer.

You can check each solution algebraically by substituting each solution in
the original equation. If, after the substitution, the left side of the
original equation equals the right side of the original equation, the
solution is valid.

You can also check the solutions graphically by graphing the function formed
by subtracting the right side of the original equation from the left side of
the original equation. The solutions of the original equation are the
x-intercepts of this graph.

Algebraic Check:

Check solution x=0

Left Side:

Right Side:

Since the left side of the original equation equals the right side of the
original equation when you substitute 0 for x, then 0 is a solution.

Check solution
x=1.04719755

Left Side:

Right Side:

Since the left side of the original equation equals the right side of the
original equation when you substitute
1.04719755 for x, then
1.04719755is a solution.

Check solution
x=-1.04719755

Left Side:

Right Side:

Since the left side of the original equation equals the right side of the
original equation when you substitute
-1.04719755 for x, then
-1.04719755is a solution.

We have just verified algebraically that the exact solutions are x=0 and
and these solutions repeat
every
units. The approximate values of these solutions are
and
and these solutions repeat every
units.

Graphical Check:

Graph the equation
Note that the graph crosses the x-axis many times indicating many solutions.
Let's check a few of these x-intercepts against the solutions we
derived.

The graph crosses the x-axis at 0. Since the period is
,
you can verify that the graph also crosses the x-axis again at
6.2831853 and at
,
etc.

The graph crosses the x-axis at
Since the period is
,
the graph also crosses the x-axis again at
1.04719755+6.2831853=7.3303829 and at
,
etc.

The graph crosses the x-axis at
.
Since the period is
,
the graph also crosses the x-axis again at
-1.04719755+6.2831853=5.235988 and at
,
etc.

Note: If the problem were to find the solutions in the interval
,
then you choose those solutions from the set of infinite
solutions that belong to the set
,
5.235988, and